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प्रश्न
A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price is given as p = 120 – x. Find the value of x for which revenue is increasing.
उत्तर
Let C be the total cost function.
∴ C = 40 + 2x
Revenue = Price × Demand
∴ `"R" = "p" × "x" = (120 - "x") * "x"`
∴ R = 120x - x2
∴ `"dR"/"dx" = 120 - 2"x" = 2(60 - "x")`
Since revenue R is an increasing function, `"dR"/"dx" > 0`
∴ 2(60 - x) > 0
∴ 60 - x > 0
∴ 60 > x
∴ x < 60
∴ The revenue R is increasing for x < 60.
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संबंधित प्रश्न
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Fill in the blank:
A road of 108 m length is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are _______.
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If the elasticity of demand η = 1, then demand is ______.
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Complete the following activity to find MPC, MPS, APC and APS, if the expenditure Ec of a person with income I is given as:
Ec = (0.0003)I2 + (0.075)I2
when I = 1000
If f(x) = x3 – 3x2 + 3x – 100, x ∈ R then f"(x) is ______.
If 0 < η < 1 then the demand is ______.
In a factory, for production of Q articles, standing charges are ₹500, labour charges are ₹700 and processing charges are 50Q. The price of an article is 1700 - 3Q. Complete the following activity to find the values of Q for which the profit is increasing.
Solution: Let C be the cost of production of Q articles.
Then C = standing charges + labour charges + processing charges
∴ C = `square`
Revenue R = P·Q = (1700 - 3Q)Q = 1700Q- 3Q2
Profit `pi = R - C = square`
Differentiating w.r.t. Q, we get
`(dpi)/(dQ) = square`
If profit is increasing , then `(dpi)/(dQ) >0`
∴ `Q < square`
Hence, profit is increasing for `Q < square`
